Can someone help me solve this equation? Step by Step.

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Can someone help me solve this equation? Step by Step.

Mathematics
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HI!!
is it a hard equation ?
\[\sqrt{x+6}-4=x\]

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oh no that can't be that hard
Can you help me? I've never had this type of equation equaled to x and it sort of makes me nervous. Can you walk me through it? @misty1212
yeah no problem first step is to get the radical by itself on one side of the equation by adding 4 to both sides, i.e. write \[\sqrt{x+6}=x+4\]
now you want to do it the math teacher way , or you want to think what the answer is , or both?
I'm not sure I understand what you're asking.
we can find the answer in our head, but that is probably not what your math teacher wants you to do ,so lets go ahead and solve it the way they do in math class
Okay
i can show you how to get the answer in your head second
Alright thanks :)
you want to get rid of the radical, so square both sides (carefully)
How would I square (x+4). Like this? x^2+8
\[\sqrt{x+6}=x+4\] square and get \[x+6=(x+4)^2\] or \[x+6=(x+4)(x+4)\] or \[x+6=x^2+8x+16\]
Where did the 8x come from?
no dear that is not how you square \(x+4\) you have to write it as \[x(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16\]
it is that old "first outer inner last" stuff
Ohhhh, FOIL okay..that makes sense.
so now we have \[x+6=x^2+8x+16\] which is a quadratic equation we have to solve
set it equal to zero, get \[x^2+7x+10=0\] then factor
\[x^2+7x+10=0\\ (x+2)(x+5)=0\] so \[x=-2,x=-5\] but we are not done yet
you have to check the answer in the original equation not the quadratic equation if you do, you will see that \(x=-2\) works but \(x=-5\) does not
Making it an extraneous solution correct?
right , which is a silly way of saying it is not a solution but no matter
:) thank you @misty1212
\[\color\magenta\heartsuit\]

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